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Ronch [10]
3 years ago
11

An office building is served by an air-cooled chiller currently operating at 115 tons (404.5 kW). The measured chilled water sup

ply temperature is 43°F (6.1°C), and the measured chilled water return temperature is 57°F (13.9°C). What is the estimated chilled water flow rate through the chiller? A. 172 gpm (10.9 L/s) B. 197 gpm (12.4 L/s) C. 212 gpm (13.4 L/s)
Engineering
1 answer:
Andrei [34K]3 years ago
8 0

Answer:

B.197 gpm and 12.4 L/s

Explanation:

Given that

Load Q = 404.5 KW

Water inlet temperature= 6.1 °C

Water outlet temperature= 13.9°C

We know that specific heat for water

C_p=4.187\ \frac{KJ}{kg.K}

Now from energy balance

Q=\dot{m}C_p\Delta T

by putting the values

Q=\dot{m}C_p\Delta T

404.5=\dot{m}\times 4.187(13.9-6.1)

\dot{m}=12.38\ \frac{kg}{s}     (1 Kg/s = 15.85 gal/min)

We can say that

\dot{m}=196.31\ \frac{gal}{min}

We know that

\dot{m}=\rho\times volume\ flow\ rate

12.38=1000 x volume flow rate

volume\ flow\ rate\ = 12.38\times 10^{-3}\ \frac{m^3}{s}

So

volume flow rate = 12.38 L/s

So the option B is correct.

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An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft.
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Answer:

a. L_o  = 40 psf

b. L ≈ 30.80 psf

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d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, L_o  = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)

For the school, K_{LL} = 2

Therefore, we have;

L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, W_d = b × W_{D + L} =

∴  W_d =  = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, W_d = 812.8 ft./lb

d. For the uniformly distributed load, we have;

V_{max} = 812.8 × 26/2 = 10566.4 lbs

M_{max} =  812.8 × 26²/8 = 68,681.6 ft-lbs

v_{max} = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

P_L = 1000 lb

W_{D} = 20 × 16 = 320 lb/ft.

V_{max} = 1,000 + 320 × 26/2 = 5,160 lbs

M_{max} =  1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

v_{max} = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

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A thin plastic membrane is used to separate helium from a gas stream. Under steady-state conditions the concentration of helium
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N_A=1.5*10^-8 kmol/s.m^2

Explanation:

<u>KNOWN: </u>

Molar concentration of helium at the inner and outer surfaces of a plastic membrane. Diffusion coefficient and membrane thickness.  

<u>FIND:</u>

Molar diffusion flux.  

<u>ASSUMPTIONS:</u>

(1) Steady-state conditions, (2) One-dimensional diffusion in a plane wall, (3) Stationary medium, (4) Uniform C = C_A + C_B.  

<u>ANALYSIS:</u> The molar flux may be obtained from

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N_A=1.5*10^-8 kmol/s.m^2

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